Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $t \neq 0$. $r = \dfrac{-2t - 14}{t - 10} \times \dfrac{t - 10}{t^2 - 3t - 70} $
Solution: First factor the quadratic. $r = \dfrac{-2t - 14}{t - 10} \times \dfrac{t - 10}{(t + 7)(t - 10)} $ Then factor out any other terms. $r = \dfrac{-2(t + 7)}{t - 10} \times \dfrac{t - 10}{(t + 7)(t - 10)} $ Then multiply the two numerators and multiply the two denominators. $r = \dfrac{ -2(t + 7) \times (t - 10) } { (t - 10) \times (t + 7)(t - 10) } $ $r = \dfrac{ -2(t + 7)(t - 10)}{ (t - 10)(t + 7)(t - 10)} $ Notice that $(t - 10)$ and $(t + 7)$ appear in both the numerator and denominator so we can cancel them. $r = \dfrac{ -2\cancel{(t + 7)}(t - 10)}{ (t - 10)\cancel{(t + 7)}(t - 10)} $ We are dividing by $t + 7$ , so $t + 7 \neq 0$ Therefore, $t \neq -7$ $r = \dfrac{ -2\cancel{(t + 7)}\cancel{(t - 10)}}{ (t - 10)\cancel{(t + 7)}\cancel{(t - 10)}} $ We are dividing by $t - 10$ , so $t - 10 \neq 0$ Therefore, $t \neq 10$ $r = \dfrac{-2}{t - 10} ; \space t \neq -7 ; \space t \neq 10 $